There's an asymmetry between the requirements for intersections and unions in the laws for topological spaces. In the most common presentation of a topological space that I've seen, the space is a pair $(X, \tau)$ with $X$ being the "universe of discourse" (I don't know what the actual term is), and $\tau$ being the set of open sets. I think there's an equivalent one with closed sets that you get by DeMorgan-ing the laws for open sets.
Anyway, I've studied mathematical objects before where you have a set and you insist that it's closed under certain operations. With topological spaces, you have two different closure properties. I think you can in general "translate" between closure-flavored and fixpoint-flavored properties... something like
$$ \text{S is closed under f} \Longleftrightarrow \text{S is a fixed point of } (\lambda x : x \cup \left\{f(t)\;|\;t \in x \right\}) $$
How do you do the same thing for a topological space?
Pretend for a moment that all topological spaces that we consider are topological spaces over a particular set $X$.
The laws for topological spaces ($\tau \subseteq \mathcal{P}(X) $ or equivalently $\tau \in \mathcal{P}(\mathcal{P}(X))$) are the following. (I'm not sure whether "law" is the right word here.)
- $\varnothing \in \tau$
- $X \in \tau$
- let $A$ be a finite sequence of elements of $\tau$ of length $n$ $$ \left( \bigcap_{i=1}^n A_i \right) \in \tau $$
- let $B \in \mathcal{P}(\tau)$ be an arbitrary element of the powerset of $\tau$. $$ \left( \bigcup B \right) \in \tau $$
suppose for a second that we ignore law #2, so we're talking about topological spaces on $X$ or on some subset of $X$.
I want to come up with a function $T : \mathcal{P}(\mathcal{P}(X)) \rightarrow \mathcal{P}(\mathcal{P}(X))$ so that all the fixed points are topological spaces on $X$ (or a subset of $X$).
Here's what I came up with
$$ T : \mathcal{P}(\mathcal{P}(X)) \rightarrow \mathcal{P}(\mathcal{P}(X))$$ $$ T(\tau) = \left\{ (\cap \; x) \; | \; x \in \mathcal{P}(\tau), \text{x is finite}\right\} \cup \left\{ (\cup \; x) \; | \; x \in \mathcal{P}(\tau) \right\} $$
It's possible to "factor" this function into two functions, since $\psi \subseteq T(\psi)$ for every possible set $\psi$.
$$ T = T_1 \circ T_0 $$
where
$$ T_0(\tau) = \left\{ (\cap \; x) \; | \; x \in \mathcal{P}(\tau), \text{x is finite} \right\} $$
and
$$ T_1(\tau) = \left\{ (\cup \; x) \; | \; x \in \mathcal{P}(\tau)\right\} $$
Suppose I have a function $T'$ $$ T' = \left\{ f(x,x') \; | \; (x,x') \in \tau \times \tau \right\}$$
In $T'_1$, $f$ is $\cup$. In $T'_0$, $f$ is $\cap$.
Can $T_0$ be replaced with $T'_0$?
Can $T_1$ be replaced with $T'_1$?
I know that both can't be true simultaneously...
If a family of sets is closed under the map $a, b\mapsto a\cap b$ (that is, individual intersections), then it's easy to show by induction that it's closed under all finite intersections. So $T_0'$ can indeed replace $T_0$.
However, this would not work for infinite intersections. And indeed, $T_1$ cannot be replaced by $T_1'$. For an easy example, start with the set of all finite subsets of $\mathbb{N}$. This is closed under $T_1'$, but not under $T_1$, since $\mathbb{N}$ is a union of infinitely many finite subsets of $\mathbb{N}$ but is not a union of finitely many finite subsets of $\mathbb{N}$.